Determine the acceleration of the block of mass M in the...

Let $a$ be the acceleration of block $M$ towards the right.

Let $x$ be the distance of block $M$ from the wall and $y$ be the distance of block $m$ below the pulley on $M$. The total length of the string is $L = 2x + y + \text{constant}$.

Differentiating twice with respect to time, we get $0 = -2a + a_{y,\text{rel}}$, which implies the downward acceleration of $m$ relative to $M$ is $a_{y,\text{rel}} = 2a$.

The absolute horizontal acceleration of $m$ is $a$ (towards the right), and its absolute vertical acceleration is $2a$ (downwards).

For block $m$ in the horizontal direction, the only force is the normal reaction $N$ from $M$:
$N = ma$

For block $m$ in the vertical direction, the forces are gravity $mg$ (downwards), tension $T$ (upwards), and friction $f_1$ (upwards):
$mg - T - f_1 = m(2a)$

Since $f_1 = \mu_1 N = \mu_1 ma$, we have:
$T = mg - 2ma - \mu_1 ma = m[g - a(2 + \mu_1)]$

For the system of blocks $(M+m)$ in the vertical direction, the internal forces (tension $T$ and friction $f_1$ between $M$ and $m$) cancel out. The net vertical acceleration of the center of mass is due to the downward acceleration of $m$.
The normal reaction from the ground is $N_{\text{ground}}$.
$(M+m)g - N_{\text{ground}} = m(2a) + M(0)$
$N_{\text{ground}} = (M+m)g - 2ma$

The friction force from the ground on $M$ is:
$f_2 = \mu_2 N_{\text{ground}} = \mu_2 [(M+m)g - 2ma]$

For the system $(M+m)$ in the horizontal direction, the external forces are the tensions from the two horizontal string segments pulling to the right, and the ground friction $f_2$ pulling to the left:
$2T - f_2 = (M+m)a$

Substituting the expressions for $T$ and $f_2$ into the horizontal equation:
$2m[g - a(2 + \mu_1)] - \mu_2 [(M+m)g - 2ma] = (M+m)a$

Expanding and rearranging the terms:
$2mg - 4ma - 2\mu_1 ma - \mu_2(M+m)g + 2\mu_2 ma = Ma + ma$

$2mg - \mu_2(M+m)g = Ma + 5ma + 2\mu_1 ma - 2\mu_2 ma$

$g[2m - \mu_2(M+m)] = a[M + m(5 + 2\mu_1 - 2\mu_2)]$

$a = \dfrac{[2m - \mu_2(M+m)]g}{M + m[5 + 2(\mu_1 - \mu_2)]}$

Answer: $\dfrac{[2m - \mu_2(M+m)]g}{M + m[5 + 2(\mu_1 - \mu_2)]}$